gcd

To find the greatest common divisor (GCD) of $2^{1001}-1$ and $2^2-1$, we can use the difference of squares factorization.

Notice that $2^2 - 1$ is a difference of squares: $2^2 - 1 = (2+1)(2-1) = 3$.

Now, let's use the difference of squares formula for $2^{1001}-1$:

$$2^{1001} - 1 = (2^{500} + 1)(2^{500} - 1).$$

Again, $2^{500} - 1$ is another difference of squares: $2^{500} - 1 = (2^{250} + 1)(2^{250} - 1)$.

This pattern continues, and we eventually reach $2^2 - 1 = 3$.

So, we have:

$$2^{1001} - 1 = (2^{500} + 1)(2^{500} - 1) = (2^{250} + 1)(2^{250} - 1) = \dots = 3^8.$$

Now, we can see that the greatest common divisor of $2^{1001}-1$ and $2^2-1$ is simply $3$.

Therefore, the greatest common divisor of $2^{1001}-1$ and $2^2-1$ is $\boxed{3}$.

Partial factorization of:  2^(1001) - 1
23×89×127×911×6007×8191×724153×112901153×23140471537×158822951431×6120360210855167691724912383945435257223721665116428926703276245201747560818857289386060189570105107203570707721973897692732984143624435639152229768723838829356129391428019897940054854494968142136307334184139636418088074652883916814682909620534381239
(10 prime factors, 1 composite factor)
2^2 - 1 ==3
Prime factorization of:
3 ==1 x 3
Since there no common factor between 2^(1001) - 1 and 2^2 - 1, therefore:

The GCD of [2^(1001) - 1,  2^2 - 1]==1